ANSI Common Lisp. Why I get an other answer in the last case?
(list 1 2 3 nil) ; (1 2 3 nil)
(funcall (function list) 1 2 3 nil) ; (1 2 3 nil)
(apply (function list) '(1 2 3 nil)) ; (1 2 3 nil)
(apply (function list) 1 2 3 nil) ; (1 2 3)
APPLY expects as arguments:
The function will basically be called with the result of (list* 0-arg ... n-arg argument-list)
Note that (list* '(1 2 3)) evaluates to just (1 2 3).
The arguments are called spreadable argument list in Common Lisp.
CL-USER 60 > (apply (function list) 1 2 3 nil)
(1 2 3)
CL-USER 61 > (apply (function list) (list* 1 2 3 nil))
(1 2 3)
CL-USER 62 > (apply (function list) (list* '(1 2 3)))
(1 2 3)
APPLY uses such a spreadable argument list by design. For example (... 1 2 3 '(4 5)). With FUNCALL you have to write the arguments as usual: (... 1 2 3 4 5).
APPLY has a single purpose in Common Lisp: it allows functions to be called with computed argument lists. To make that a bit more convenient, this idea of the spreadable argument list has been used. It works the same for example in Emacs Lisp.
Imagine that you have a list of arguments and you want to add two arguments in front.
CL-USER 64 > (apply '+ args)
60
CL-USER 65 > (apply '+ 1 2 args)
63
CL-USER 66 > (apply '+ (append (list 1 2) args))
63